d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)

e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)

c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)

\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)

\(\Leftrightarrow x-4=0\)

hay x=4

a) \(\sqrt{x-1+2\sqrt{x-1}.1+1^2}=2;đk:x\)≥1
⇔\(\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}.1+1^2}=2\left(hđt-1\right)\)
⇔\(\sqrt{\left(\sqrt{x-1}+1\right)^2=2}\)
⇔|\(\sqrt{x-1}+1\)|=2
⇔\(\left[{}\begin{matrix}\sqrt{x+1}-1=2\\\sqrt{x+1-1}=-2\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\sqrt{x+1}=3\\\sqrt{x+1}=-1\left(L\right)\end{matrix}\right.\)⇔x+1=9⇔x=10(TM)
→S={10}

a) `sqrt(x^2-6x _9) = 4-x`

`<=> sqrt[(x-3)^2] =4-x`

`<=> |x-3| =4-x ( đk :x<=4)`

`<=> |x-3| = |4-x|`

`<=> [(x-3 =4-x),(x-3 = x-4):}`

`<=>[(x = 7/2(t//m)),(0=-1(vl)):}`

Vậy `S = {7/2}`

b) `sqrt(x^2 -9) + sqrt(x^2 -6x +9) =0(đk : x>=3(hoặc) x<=-3)`

`<=>sqrt(x^2 -9) =- sqrt(x^2 -6x +9) `

`<=>(sqrt(x^2 -9))^2 =(- sqrt(x^2 -6x +9))^2`

`<=> x^2 -9 = x^2 -6x +9`

`<=> 6x = 9+9 =18`

`<=> x=3(t//m)`

Vậy `S={3}`

c) `sqrt(x^2 -2x+1) + sqrt(x^2-4x+4) =3`

`<=> sqrt[(x-1)^2] +sqrt[(x-2)^2] =3`

`<=> |x-1| +|x-2| =3`

xét `x<1 =>{(|x-1| =1-x ),(|x-2|=2-x):}`

`=> 1-x +2-x =3`

`=> x = 0(t//m)`

xét `1<=x<2 => {(|x-1|=x-1),(|x-2|= 2-x):}`

`=> x-1 +2-x =3`

`=>1=3 (vl)`

xét `x>=2 => {(|x-1| =x-1),(|x-2|=x-2):}`

`=> x-1+x-2 =3`

`=> x=3(t//m)`

Vậy `S = {0;3}`

a: =>|x-3|=4-x

TH1: x>=3

=>4-x=x-3

=>x=7/2(nhận)

TH2: x<3

=>3-x=4-x(loại)

b: =>căn x-3(căn x+3+căn x-3)=0

=>x-3=0

=>x=3

c: =>|x-1|+|x-2|=3

Th1: x<1

=>1-x+2-x=3

=>x=0(nhận)

TH2: 1<=x<2

=>x-1+2-x=3

=>1=3(loại)

TH3: x>=2

=>x-1+x-2=3

=>x=3

a. ĐKXĐ: $x\geq 2$ hoặc $x=1$

PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$

$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)

b.

PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$

$\Leftrightarrow |x-2|=|2x-3|$

\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)

c. ĐKXĐ: $x=2$ hoặc $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)

d.

PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$

$\Leftrightarrow |2x-1|=|x-3|$

\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)

a: Ta có: \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)

\(\Leftrightarrow x^2-3x+2=x-1\)

\(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=-x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)

c: Ta có: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)

\(\Leftrightarrow x^2-5x+6=x-2\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)

TH₁: \(x\ge3\)

\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)

TH₂: \(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH₃: \(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH₄: \(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)

Lời giải:

a. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$

$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$

$\Leftrightarrow |\sqrt{x-4}+2|=2$

$\Leftrightarrow  \sqrt{x-4}+2=2$

$\Leftrightarrow \sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

b. ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$

$\Leftrightarrow |2x-1|=|x-3|$

\(\Rightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)

c.

PT \(\Rightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x(x-1)=0\end{matrix}\right.\Rightarrow x=1\)

Đăng 1 lúc mà nhiều thế. Lần sau đăng 1 câu thôi b.

b/ \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)

Ta có: \(VT\ge1+2+\sqrt{5}=3+\sqrt{5}\)

Dấu = xảy ra khi \(x=2\)

c/ \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=\sqrt{3-\left(x-1\right)^2}+\sqrt{1-\left(x+3\right)^2}\)

\(\le1+\sqrt{3}\)

Dấu = không xảy ra nên pt vô nghiệm

Câu d làm tương tự

\(a,\sqrt{x^2-4}-x^2+4=0\) 

\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\) 

\(\Leftrightarrow x^2-4=\left(x-4\right)^2\) 

\(\Leftrightarrow x^2-4-x^4+8x^2-16=0\)  

\(\Leftrightarrow-x^4-7x^2-20=0\) 

\(\Leftrightarrow-\left(x^4+7x^2+\frac{49}{4}\right)-\frac{31}{4}=0\) 

\(\Leftrightarrow-\left(x^2+\frac{7}{2}\right)^2=\frac{31}{4}\) 

\(\Leftrightarrow\left(x^2+\frac{7}{2}\right)=-\frac{31}{4}\) 

\(\Rightarrow\)pt vô nghiệm

a/ĐK: \(x\ge2\)

 \(PT\Leftrightarrow x^2-4=\sqrt{x^2-4}\)

Đặt \(x^2-4=t\Rightarrow x^2=t+4\)

Thay vào,phương trình đã cho tương đương với: 

\(t=\sqrt{t}\Leftrightarrow t^2=t\Rightarrow\orbr{\begin{cases}t=1\\t=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-4=1\\x^2-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=5\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}\\x=2\end{cases}}\)  (t/m)

a: ĐKXĐ: x>=-3/2

\(\sqrt{x^2+4}=\sqrt{2x+3}\)

=>\(x^2+4=2x+3\)

=>\(x^2-2x+1=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1(nhận)

b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))

=>\(\sqrt{\left(x-3\right)^2}=2x-1\)

=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>x=4/3(nhận) hoặc x=-2(loại)

c:

Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)

ĐKXĐ: \(x>=-3\)

\(\sqrt{4x+12}=\sqrt{9x+27}-5\)

=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)

=>\(-\sqrt{x+3}=-5\)

=>x+3=25

=>x=22(nhận)

d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)

=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)

=>\(4x^2-6x+1=4x^2-20x+25\)

=>\(-6x+20x=25-1\)

=>\(14x=24\)

=>x=12/7(nhận)

Đề yc giải pt à em?

Câu b bạn có bị lỗi dấu căn không mà sao nó kéo dài cả 2 vế pt vậy :v

\(a,\sqrt{x^2-6x+9}+x=11\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=11-x\)

\(\Leftrightarrow\left|x-3\right|=11-x\\ TH_1:x\ge3\\ x-3=11-x\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\left(tm\right)\)

\(TH_2:x< 3\\ -x+3=11-x\\ \Leftrightarrow-x+x=11-3\\ \Leftrightarrow0=8\left(VL\right)\)

Vậy \(S=\left\{7\right\}\)

\(c,\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\) \(\left(dk:x\ge-1\right)\)

\(\Leftrightarrow\sqrt{4^2}.\sqrt{\left(x+1\right)}-\sqrt{3^2}.\sqrt{\left(x+1\right)}=4\left(1\right)\)

Đặt \(a=\sqrt{x+1}\left(a\ge0\right)\)

Pt trở thành : \(4a-3a=4\Leftrightarrow a=4\left(tmdk\right)\)

\(\Rightarrow\sqrt{x+1}=4\\ \Rightarrow\left(\sqrt{x+1}\right)^2=16\\ \Rightarrow\left|x+1\right|=16\)

\(TH_1:x\ge-1\\ x+1=16\Leftrightarrow x=15\left(tm\right)\\ TH_2:x< -1\\ -x-1=16\Leftrightarrow x=-17\left(tm\right)\)

Nhưng loại TH2 vì dk ban đầu là \(x\ge-1\)

Vậy \(S=\left\{15\right\}\)

\(d,\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\left(dk:x\ge-1\right)\\ \Leftrightarrow\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}-\sqrt{x+1}=0\)

Đặt \(\sqrt{x+1}=a\left(a\ge0\right)\)

Tới đây bạn làm tương tự câu c nha.